Posted by Sarah on Tuesday, February 10, 2009 at 10:29pm.
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 1.5 x 107 C/m2, and the plates are separated by a distance of 1.4 x 102 m. How fast is the electron moving just before it reaches the positive plate?
THANK YOU for any help or hints or anything that you can give me!

Physics. Speed of electron...  bobpursley, Tuesday, February 10, 2009 at 11:50pm
I would find the voltage on the capacitor, or Electric field first. Using Gauss Law, then E= sigma/2epsilion
Force/q= E
Force=Eq
force*distance=Eq*distance
1/2 m v^2=Eq*distance
solve for velocity v. 
Physics. Speed of electron...  drwls, Tuesday, February 10, 2009 at 11:51pm
The E field between the plates is proportional to the charge per unit area, sigma (1.5 x 107 C/m2).
E = sigma/epsilon
where epsilon is the permittivity of free space, which you should look up.
The kinetic energy that is given to the electrion while crossing the gap of length X is
KE = e E X
Set that equal to the final kinetic energy and solve for the velocity. 
Physics. Speed of electron...  bobpursley, Tuesday, February 10, 2009 at 11:57pm
Dr WLS is correct, E between the plates is sigma/episilon, not sigma/2epsilon.